1.A.) (25 POINTS) Decrypt the ciphertext C by using brute force attack. Please show your
1.B.) (25 POINTS) Decrypt the ciphertext C by using the φ(i) correlation model explained in
Following lists the monogram English statistics where p(c) represents the frequency of
character c in English texts.
c p(c) c p(c) c p(c) c p(c)
Figure 1. Monogram English Statistics.
Let φ(i) be the correlation of the frequency of each letter in the ciphertext with the character
frequencies in English obtained from Figure 1. Let f(c) be the frequency of character c
(expressed as a fraction). The formula for the correlation for each ciphertext (with all
arithmetic being mod 26) is
φ(i) = Σ0 ≤ c ≤ 25 f(c) x p(c – i)
where f(c) is the ciphertext character frequency and p(c-i) is obtained from Figure 1. This
correlation φ(i) should be a maximum when the key k translates the ciphertext into
English. Trying the most likely key first, i.e. the i value that yields the largest φ(i), we obtain
plaintext alternatives. Out of these, the alternative that is a meaningful English text is decided
to be the plaintext.
As an example, assume that ciphertext “KHOOR ZRUOG” is given. To solve it using the φ(i)
model, we first compute the frequency (as a fraction) of each letter in the ciphertext as
where ciphertext frequencies sum up to 1.0. Then, we compute φ(i) as follows:
φ(i) = Σ0 ≤ c ≤ 25 f(c) x p(c – i) = 0.1 x p(6 – i) + 0.1 x p(7 – i) + 0.1 x p(10 – i) + 0.3 x p(14 – i) +
0.2 x p(17 – i) + 0.1 x p(20 – i) + 0.1p(25 – i)
For example, for i=0, the φ(i) is computed as:
φ(0) = 0.1 x p(6) + 0.1 x p(7) + 0.1 x p(10) + 0.3 x p(14) + 0.2 x p(17) + 0.1 x p(20) + 0.1 x p(25)
= 0.1 x 0.015 + 0.1 x 0.06 + 0.1 x 0.005 + 0.3 x 0.08 + 0.2 x 0.065 + 0.1 x 0.03 + 0.1 x 0.002
Therefore, we obtain the table below:
Figure 2. The value of φ(i) for 0 ≤ i ≤ 25
The most likely keys (represented as i) that yield the largest φ(i) value are highlighted in
Figure 2: They are 6, 10, 3, and 14 in decreasing order of φ(i) value.
Trying the most likely key first, we obtain as plaintext “EBIIL TLOIA” when i = 6, “AXEEH
PHKEW” when i = 10, “HELLO WORLD” when i = 3, and “WTAAD LDGAS” when i = 14.
The statistics indicated that the key was most likely 6, when in fact the correct key was 3. So
the attacker must test the results. The statistics simply reduce the number of trials in most
cases. Only three trials were needed.
IMPORTANT NOTE: You may either write computer program or work manually to solve
this question. It is suggested to use a program as it will reduce the amount of time you spend
on it significantly. If writing a code, please submit your source code together with screenshots.
If solved manually, please provide your solution steps clearly.
2.) (50 POINTS + 20 EXTRA POINTS If you use an alternative method to find the
multiplicative inverse of a matrix other than what is described in this question) Hill cipher is a
well known polyalphabetic cipher. In a hill cipher cryptosystem, the encryption and
decryption formulas are given as follows:
Encryption: C = (P x K ) mod S
Decryption: P = (C x K-1
) mod S
where P is plaintext, C is ciphertext, x is matrix multiplication, key K is a matrix of size a x a,
S is the source alphabet size (which is 26 for English), and K-1
is the multiplicative inverse of
the matrix K such that:
(K x K-1
) mod S = I
where I is the identity matrix. For encryption and decryption, numeric correspondent of each
letter (0 for letter a, 1 for letter b, .., 25 for letter z) is used. We assume that only lowercase
letters are used as plaintext P letters. Given the key matrix as:
17 17 5
K = 21 18 21
2 2 19
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